Question

a differential equation is an equation involving an unknown function and its derivatives. consider the differential equation y(t)+y(t)=0.
a. show that y = a sin t satisfies the equation for any constant a.
b. show that y = b cos t satisfies the equation for any constant b.
c. show that y = a sin t + b cos t satisfies the equation for any constants a and b.
a. find y(t) for y = a sin t. choose the correct answer below.
a. y(t)=a cos t
b. y(t)= - a cos t
c. y(t)=a sin t
d. y(t)= - a sin t
does this show that y(t)+y(t)=0 for y = a sin t?
no
yes

Explanation:

Step1: Find first - derivative of $y = A\sin t$

Using the derivative formula $\frac{d}{dt}(\sin t)=\cos t$, we have $y'(t)=A\cos t$.

Step2: Find second - derivative of $y = A\sin t$

Differentiate $y'(t)=A\cos t$ with respect to $t$. Since $\frac{d}{dt}(\cos t)=-\sin t$, we get $y''(t)=-A\sin t$.

Step3: Substitute $y$ and $y''$ into the differential equation

Substitute $y = A\sin t$ and $y''(t)=-A\sin t$ into $y''(t)+y(t)$. We have $y''(t)+y(t)=-A\sin t + A\sin t=0$.

For part b:

Step1: Find first - derivative of $y = B\cos t$

Using the derivative formula $\frac{d}{dt}(\cos t)=-\sin t$, we get $y'(t)=-B\sin t$.

Step2: Find second - derivative of $y = B\cos t$

Differentiate $y'(t)=-B\sin t$ with respect to $t$. Since $\frac{d}{dt}(\sin t)=\cos t$, we have $y''(t)=-B\cos t$.

Step3: Substitute $y$ and $y''$ into the differential equation

Substitute $y = B\cos t$ and $y''(t)=-B\cos t$ into $y''(t)+y(t)$. We get $y''(t)+y(t)=-B\cos t + B\cos t = 0$.

For part c:

Step1: Find first - derivative of $y=A\sin t + B\cos t$

Using the sum - rule of differentiation and the derivative formulas $\frac{d}{dt}(\sin t)=\cos t$ and $\frac{d}{dt}(\cos t)=-\sin t$, we have $y'(t)=A\cos t - B\sin t$.

Step2: Find second - derivative of $y=A\sin t + B\cos t$

Differentiate $y'(t)=A\cos t - B\sin t$ with respect to $t$. We get $y''(t)=-A\sin t - B\cos t$.

Step3: Substitute $y$ and $y''$ into the differential equation

Substitute $y = A\sin t + B\cos t$ and $y''(t)=-A\sin t - B\cos t$ into $y''(t)+y(t)$. We have $y''(t)+y(t)=(-A\sin t - B\cos t)+(A\sin t + B\cos t)=0$.

Answer:

a. The correct answer for $y''(t)$ when $y = A\sin t$ is D. $y''(t)=-A\sin t$, and $y''(t)+y(t)=0$ for $y = A\sin t$ (Yes).
b. For $y = B\cos t$, $y'(t)=-B\sin t$, $y''(t)=-B\cos t$, and $y''(t)+y(t)=0$.
c. For $y = A\sin t + B\cos t$, $y'(t)=A\cos t - B\sin t$, $y''(t)=-A\sin t - B\cos t$, and $y''(t)+y(t)=0$.